Chapter 10.4, Problem 73SSC

Interpretation Introduction

Interpretation:

The ore of iron with greater percent of iron per kilogram needs to be determined.

Concept introduction:

Empirical formula is the simplest formula of any organic or inorganic compound that represents the simple ratio of all atoms present in the molecule. The empirical and molecular formula can be used to calculate the mass percent of each element given in the formula. The mathematical expression for mass percent of element can be shown as:

$\text{Mass percent = }\frac{\text{number of element ×molar mass of element }}{\text{molar mass of compound}}\times 100$

Answer to Problem 73SSC

${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ has greater percent of iron as it is 72.3 %.

Explanation of Solution

Given information:

Hematite ( ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ ) and magnetite ( ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ ) are two ores of iron.

Molar mass of Fe = 55.8 g/mol

Molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ = 159.7 g/mol

Molar mass of ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ = 231.5 g/mol

$\text{Mass percent = }\frac{\text{number of element ×molar mass of element }}{\text{molar mass of compound}}\times 100$

Calculate mass percent of Fe:

$\begin{array}{l}{\text{In Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{:}\\ \text{Mass percent of Fe = }\frac{\text{2 × 55}\text{.8 g/mol }}{\text{159}\text{.7 g/mol }}\text{×100 = 69}\text{.9\%}\\ {\text{In Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{:}\\ \text{Mass percent of Fe = }\frac{\text{3 × 55}\text{.8 g/mol }}{\text{231}\text{.5 g/mol }}\text{×100 = 72}\text{.3 \%}\end{array}$

Thus ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ has greater percent of iron as it is 72.3 %.

Conclusion

${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ has greater percent of iron as it is 72.3 %.